I have the following:
Consider the real vectorspace with bounded functions
$$V = \{f:[0,1]\rightarrow\mathbb{R} | \exists C > 0 : f([0,1])\subset[-C,C]\}$$
and the supremum-norm
$$||f||_\infty = sup \{|f(x)| | x \in [0,1]\}$$
Prove that $(V, ||\cdot||_\infty)$ is complete metric space and no hilbert space.
Well, I've already proven that it is a metric space. For proving that it is complete I thought about the following:
Let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy-Sequence in $V$. That means $x_n \subset [-C,C]$. That means that there is an $\epsilon > 0$ so that an $N \in \mathbb{N}$ exists with
$$d(x_n,x_m) = ||x_n - x_m||_\infty = sup\{|x_n - x_m|\} < \epsilon$$
But at this point I see that something my be wrong. First of all that sequence I defined is in $V$. That means it can only be at $[0,1]$ and thus $n = 1,2$, right? Maybe I should consider functions in $V$ instead? Could I treat a function the same way I would treat that series? Saying $||f(x_m) - f(x_n)||_\infty < \epsilon$ with $x_m,x_n \in [0,1]$, an $N\in \mathbb{N}$ and $m,n\ge N$? The idea behind all that is to try to show that $x_n$ (or $f(x_n)$) is a null-sequence.
Besides from that I don't know what to do next here.
For the thing with the Hilbert space I thought of checking for that parallelogram-identity that would have to be satisfied:
$$||w+v||_\infty^2 + ||w-v||_\infty^2 = 2||v||_\infty^2 + 2||w||_\infty^2$$
But I couldn't think of functions in $V$ that would not fulfill this. It is because $||\cdot||_\infty$ always picks the lowest top boundary of the inserted function which are real numbers between $[-C,C]$. And I don't see this rule violated for real numbers in that range. Except for maby if a result of $sup\{|f+g|\} = 2C$ would violate any rule but I think not.
Another approach would be checking for whether a Euclidean dot-product existis in $V$, wouldn't it? The only thing that could make a problem would be to check for
$$\langle f,f \rangle = ||f||_\infty^2 = (sup\{|f(x)| | x \in[0,1]\})^2 > 0$$
which at least seems obviously true... Are there other approaches or things I've done wrong or incpmplete? I have been sitting on this since yesterday now and that is all I came up with.
I am sorry for not being able to provide that much of useful thoughts...
Thank you very much, FunkyPeanut
Use the fact that any sequence $(f_n)$ of functions in this space is uniformly bounded, and thus must a convergent subsequence $(f_{n_k})\to f\in V$
Now $\|f-f_n\|\le \|f-f_{n_k}\|+\|f_{n_k}-f_n\|\lt \epsilon/2+\epsilon/2=\epsilon$, for large enough $n$.
you can use cauchy and converging subsequence to argue convergence.
As for the parallelogram law, let $u=x^2$, $v=x$ then we have
$\|u+v\|^2+\|u-v\|^2=\|x^2+x\|^2+\|x^2-x\|^2=4+1/16$
And $2(\|u\|^2+\|v\|^2)=2(\|x^2\|^2+\|x\|^2)=4$
So they are not equal.